Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I tried to prove that if $R$ is transitive then its inverse is transitive as well. $$\begin{align} & a{{R}^{-1}}b\,\,,\,\,b{{R}^{-1}}c \\ & \Rightarrow bRa\,\,,\,\,cRb \\ & \Rightarrow cRa \\ & \Rightarrow a{{R}^{-1}}c \\ \end{align}$$

is this a correct proof (or am I completely wrong and it's not even true?)

share|improve this question
add comment

2 Answers

up vote 2 down vote accepted

A note: What you call "complementary" is conventionally called the inverse relation of $R$. (It's interesting that you got the notation right.) Wikipedia says that alternative terms are converse or transpose relation. I am not sure if "complementary" is used by anyone.

True statement and correct proof.

This statement from wikipedia article on inverse relation is relevant:

If a relation is reflexive, irreflexive, symmetric, antisymmetric, asymmetric, transitive, total, trichotomous, a partial order, total order, strict weak order, total preorder (weak order), or an equivalence relation, its inverse is too.

share|improve this answer
    
Great I confused myself and thought I might be wrong :) Thanks –  Jason Jul 30 '11 at 21:04
    
You are welcome, @Jason. When in doubt, check Wikipedia :) –  Srivatsan Jul 30 '11 at 21:06
1  
The complement of a relation $R$ is defined as the relation $S$ (on the same set) which holds iff $R$ does not hold. But I have never heard the term in the wild. –  André Nicolas Jul 31 '11 at 1:39
    
@AndreN Oh yes, thanks for pointing it out. But I sure hope OP did not have this complement in mind. –  Srivatsan Jul 31 '11 at 1:42
1  
@Srivatsan Narayanan: Luckily not, the OP's proof could not be written by someone who was thinking complement. –  André Nicolas Jul 31 '11 at 2:01
show 2 more comments

Yes, that proof looks fine to me.

share|improve this answer
    
And Thanks to you 2 :) –  Jason Jul 30 '11 at 21:04
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.